JEE Main 2025 (Online) 4th April Morning Shift | Rotational Motion Question 7 | Physics

If $\vec{L}$ and $\vec{P}$ represent the angular momentum and linear momentum respectively of a particle of mass ' $m$ ' having position vector as $\vec{r}=a(\hat{i} \cos \omega t+\hat{j} \sin \omega t)$. The direction of force is

Opposite to the direction of $\vec{L}$

Opposite to the direction of $\vec{L} \times \vec{P}$

Opposite to the direction of $\vec{r}$

Opposite to the direction of $\vec{P}$

JEE Main 2025 (Online) 3rd April Morning Shift

MCQ (Single Correct Answer)

-1

A force of 49 N acts tangentially at the highest point of a sphere (solid) of mass 20 kg , kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is

JEE Main 2025 (Online) 3rd April Morning Shift Physics - Rotational Motion Question 5 English

$$ 2.5 \mathrm{~m} / \mathrm{s}^2 $$

$$ 3.5 \mathrm{~m} / \mathrm{s}^2 $$

$$ 0.25 \mathrm{~m} / \mathrm{s}^2 $$

$$ 0.35 \mathrm{~m} / \mathrm{s}^2 $$

JEE Main 2025 (Online) 2nd April Evening Shift

MCQ (Single Correct Answer)

-1

The moment of inertia of a circular ring of mass M and diameter r about a tangential axis lying in the plane of the ring is :

$\frac{3}{8} \mathrm{Mr}^2$

$2 \mathrm{Mr}^2$

$\frac{1}{2} \mathrm{Mr}^2$

$\frac{3}{2} \mathrm{Mr}^2$

JEE Main 2025 (Online) 2nd April Morning Shift

MCQ (Single Correct Answer)

-1

Moment of inertia of a rod of mass ' M ' and length ' L ' about an axis passing through its center and normal to its length is ' $\alpha$ '. Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is :

$\alpha / 4$

$\alpha / 8$

$\alpha$

$\alpha / 2$

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